On your first trip to Planet X you happen to take along a 210 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You're curious about the acceleration due to gravity on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 26.9 cm. You then pull the mass down 7.00 cm and release it. With the stopwatch you find that 9.00 oscillations take 14.3 s. Can you now satisfy your curiosity? what is the new g

4.20m/s^2

Explanation:

The mass of 210g stretch the spring by 26.9cm, 9 oscillation took 14.3 s.

Period (T) = time of oscillation / number of oscillation = 14.3 / 9 = 1.59s

Using the formula for period of a spring;

T = 2π√ (m/k)

Divide both side by 2π

T/2π = √ (m/k)

Square both side

T^2 / (2π) ^2 = m/k

Make k subject of the formula

K = 4π^2 * m / T^2

Also using Hooke's law

Since the spring was suspended from the ceiling,

F (mg) = k * DL (where g is the new gravity in m/s^2, k is the force constant of the spring in N/m and DL is the change in length in meter

Make k subject of the formula

Mg/DL = k

Since both equation equal to K then

4π^2 * m / T^2 = mg/DL

Cancel m on both side

4π^2 * DL / T^2 = g

DL = 26.9cm = 26.9/100 m = 0.269 m

T = 1.59s and π = 3.142

Substitute these into the equation

g = (4 * 3.142 * 3.142 * 0.269) / (1.59^2) = 4.20 m/s^2